# What conditions must the constants, the b?s, satisfy so that each of these systems has a...

## Question:

What conditions must the constants, the {eq}b?s {/eq}, satisfy so that each of these systems has a solution? Hint. Apply Gauss?s Method and see what happens to the right side.

{eq}x - 3y = b_1 \\ 3x + y = b_2 \\ x + 7y = b_3 \\ 2x + 4y = b_4 {/eq}

## Gaussian Elimination

To solve a system of linear equations, we can use the Gaussian elimination, which is a technique to solve a system of linear equations by

eliminating the unknowns from the equations in such a manner that the matrix of the system is an upper triangular matrix.

The number of the non-pivot columns gives the number of free-parameters used in the infinite solution.

If the last column of the augmented matrix is a pivot column, then the system is inconsistent and there is no solution.

## Answer and Explanation:

To determine the condition of the right hand side constants, {eq}\displaystyle b {/eq}

of the system of equations {eq}\displaystyle \begin{cases} x - 3y = b_1 \\ 3x + y = b_2 \\ x + 7y = b_3 \\ 2x + 4y = b_4 \end{cases} {/eq}

we will obtain a row echelon form of the augmented matrix {eq}\displaystyle \left[\begin{array}{cc|c} 1&-3&b_1\\ 3&1&b_2\\ 1&7&b_3\\ 2&4&b_4 \end{array} \right] {/eq} as below.

{eq}\displaystyle \begin{align}&\left[\begin{array}{cc|c} 1&-3&b_1\\ 3&1&b_2\\ 1&7&b_3\\ 2&4&b_4 \end{array} \right]\overset{-3R_1+R_2}{\implies } \left[\begin{array}{cc|c} 1&-3&b_1\\ 0&10&b_2-3b_1\\ 1&7&b_3\\ 2&4&b_4 \end{array} \right]\overset{-R_1+R_3}{\implies } \left[\begin{array}{cc|c} 1&-3&b_1\\ 0&10&b_2-3b_1\\ 0&10&b_3-b_1\\ 2&4&b_4 \end{array} \right]\overset{-2R_1+R_4}{\implies } \left[\begin{array}{cc|c} 1&-3&b_1\\ 0&10&b_2-3b_1\\ 0&10&b_3-b_1\\ 0&10&b_4-2b_1 \end{array} \right]\overset{-R_2+R_3}{\implies } \left[\begin{array}{cc|c} 1&-3&b_1\\ 0&10&b_2-3b_1\\ 0&0&b_3-b_2+2b_1\\ 0&10&b_4-2b_1 \end{array} \right]\overset{-R_2+R_4}{\implies } \left[\begin{array}{cc|c} 1&-3&b_1\\ 0&10&b_2-3b_1\\ 0&0&b_3-b_2+2b_1\\ 0&0&b_4-b_2+b_1 \end{array} \right]\\\\ \implies &\boxed{\text{ for the system to be consistent and admit solutions, we need to have the following conditions } \begin{cases} b_3-b_2+2b_1=0\\ b_4-b_2+b_1=0 \end{cases}}. \end{align} {/eq}

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from High School Algebra II: Homework Help Resource

Chapter 8 / Lesson 8